Laplace-likelihood interval evidence ($Z/C=(s/w)(1-e^{-w/s})$)

If $\mathcal{L}(\theta )=C{e}^{-|\theta -\mu |/s}$, integrate the two exponential branches at $\mu$. An interval of half-width $w$ centred at $\mu$ has $Z/C=(s/w)(1-e^{-w/s})$. An interval entirely below the peak has $Z/C=s[e^{(b-\mu )/s}-e^{(a-\mu )/s}]/(b-a)$. These expressions show quantitatively how fit location and prior width combine.